Question

In a typical grinding operation, \(80%\) of the feed material passes through a sieve opening of \(4.75~\mathrm{mm}\); whereas, \(80%\) of the ground product passes through \(0.5~\mathrm{mm}\) opening. If the power required to grind \(2~\mathrm{tonnes/h}\) of the feed material is \(3.8~\mathrm{kW}\), the work index of the material is \underline{\hspace{2cm}. (rounded off to 2 decimal places)}

Hint:

For Bond’s law problems, first compute \(E=P/T\) (kWh per tonne), convert \(F_{80},P_{80}\) to \(\mu\)m, evaluate \(\left(1/\sqrt{P_{80}}-1/\sqrt{F_{80}}\right)\), then get \(W_i = E/[10(\cdot)]\).

Answer 1

Correct Answer: 6.25

Step 1: Use Bond’s law in specific-energy form.
Specific energy (per tonne) \[ E = 10\,W_i \left( \frac{1}{\sqrt{P_{80}}} - \frac{1}{\sqrt{F_{80}}} \right), \] where \(W_i\) is the work index (kWh/tonne), sizes are in \(\mu\mathrm{m}\).

Step 2: Convert sizes and compute the size term.
\(F_{80}=4.75~\mathrm{mm}=4750~\mu\mathrm{m}\), \(P_{80}=0.5~\mathrm{mm}=500~\mu\mathrm{m}\).
\[ \frac{1}{\sqrt{P_{80}}}-\frac{1}{\sqrt{F_{80}}} = \frac{1}{\sqrt{500}}-\frac{1}{\sqrt{4750}} = 0.044721 - 0.014522 = 0.030199. \]

Step 3: Find specific energy from given power and throughput.
Throughput \(T=2~\mathrm{t/h}\), power \(P=3.8~\mathrm{kW}\). \[ E=\frac{P}{T}=\frac{3.8}{2}=1.9~\mathrm{kWh/tonne}. \]

Step 4: Solve for work index \(W_i\).
\[ W_i=\frac{E}{10\left( \frac{1}{\sqrt{P_{80}}}-\frac{1}{\sqrt{F_{80}}}\right)} =\frac{1.9}{10\times 0.030199} =6.289 \approx \boxed{6.29~\mathrm{kWh/tonne}}. \]